David Kauchak CS52 – Spring 2015 Encryption David Kauchak CS52 – Spring 2015

Admin Assignment 6 4 more assignments: Assignment 7, due 11/13 5pm Assignments 9 & 10, due 12/9 11:59pm

Admin Midterm next Thursday Covers everything from 9/24 – 10/27 + some minor SML Will not have to write any assembly 2 pages of notes Review sessions next week (TBA)

Encryption What is it and why do we need it?

Encryption I like bananas

Encryption I like bananas

Encryption: a bad attempt I like bananas

Encryption: the basic idea I like bananas I like bananas decrypt message encrypt message send encrypted message

Encryption: a better approach the hawk sleeps at midnight

Encryption uses Where have you seen encryption used?

Encryption uses

Private key encryption I like bananas I like bananas decrypt message encrypt message send encrypted message

Private key encryption Any problems with this?

Private key encryption

Private key encryption ?

Public key encryption private key public key Two keys, one you make publicly available and one you keep to yourself

Public key encryption Share your public key with everyone

Public key encryption decrypt message encrypt message I like bananas I like bananas decrypt message encrypt message send encrypted message

Public key encryption I like bananas Only the person with the private key can decrypt! decrypt message

Modular arithmetic Normal arithmetic: a = b a is equal to b or a-b = 0 a b (mod n) a-b = n*k for some integer k or a = b + n*k for some integer k or a % n = b % n (where % is the mod operator) note that (mod n) is notation NOT the mod operator

Modular arithmetic Which of these statements are true? 12 5 (mod 7) 52 17 12 (mod 6) a-b = n*k for some integer k or a = b + n*k for some integer k or a % n = b % n (where % is the mod operator) 65 33 (mod 32)

Modular arithmetic Which of these statements are true? 12 5 (mod 7) 52 12-5 = 7 = 1*7 12 % 7 = 5 = 5 % 7 12 5 (mod 7) 92-52 = 40 = 4*10 92 % 10 = 2 = 52 % 20 52 92 (mod 10) 17-12 = 5 17 % 6 = 5 12 % 6 = 0 17 12 (mod 6) 65-33 = 32 = 1*32 65 % 32 = 1 = 33 % 32 65 33 (mod 32)

Modular arithmetic properties If: a b (mod n) then: a mod n b mod n (mod n) “mod”/remainder operator congruence (mod n)

Modular arithmetic properties If: a b (mod n) then: a mod n b mod n (mod n) More importantly: (a+b) mod n (a mod n) + (b mod n) (mod n) and (a*b) mod n (a mod n) * (b mod n) (mod n) What do these say?

Modular arithmetic Why talk about modular arithmetic and congruence? How is it useful? Why might it be better than normal arithmetic? We can limit the size of the numbers we’re dealing with to be at most n (if it gets larger than n at any point, we can always just take the result % n) The mod operator can be thought of as mapping a number in the range 0 … number-1

GCD What does GCD stand for?

Greatest Common Divisor gcd(a, b) is the largest positive integer that divides both numbers without a remainder gcd(25, 15) = ?

Greatest Common Divisor gcd(a, b) is the largest positive integer that divides both numbers without a remainder gcd(25, 15) = 5 25 15 25 5 1 15 5 3 1 Divisors:

Greatest Common Divisor gcd(a, b) is the largest positive integer that divides both numbers without a remainder gcd(100, 52) = ?

Greatest Common Divisor gcd(a, b) is the largest positive integer that divides both numbers without a remainder gcd(100, 52) = 4 100 52 100 50 25 20 10 5 4 2 1 52 13 4 2 1 Divisors:

Greatest Common Divisor gcd(a, b) is the largest positive integer that divides both numbers without a remainder gcd(14, 63) = ? gcd(7, 56) = ? gcd(23, 5) = ? gcd(100, 9) = ? gcd(111, 17) = ?

Greatest Common Divisor gcd(a, b) is the largest positive integer that divides both numbers without a remainder gcd(14, 63) = 7 gcd(7, 56) = 7 gcd(23, 5) = 1 gcd(100, 9) = 1 gcd(111, 17) = 1 Any observations?

Greatest Common Divisor When the gcd = 1, the two numbers share no factors/divisors in common If gcd(a,b) = 1 then a is relatively prime to b This a weaker condition than primality, since any two prime numbers are also relatively prime, but not vice versa

Greatest Common Divisor A useful property: If two numbers are relatively prime (i.e. gcd(a,b) = 1), then there exists a c such that a*c mod b = 1

RSA public key encryption Have you heard of it? What does it stand for?

RSA public key encryption RSA is one of the most popular public key encryption algorithms in use RSA = Ron Rivest, Adi Shamir and Leonard Adleman

RSA public key encryption Choose a bit-length k Security increase with the value of k, though so does computation Choose two primes p and q which can be represented with at most k bits Let n = pq and ϕ(n) = (p-1)(q-1) ϕ() is called Euler’s totient function Find d such that 0 < d < n and gcd(d,ϕ(n)) = 1 Find e such that de mod ϕ(n) = 1 Remember, we know one exists!

RSA public key encryption p: prime number q: prime number n = pq ϕ(n) = (p-1)(q-1) d: 0 < d < n and gcd(d,ϕ(n)) = 1 e: de mod ϕ(n) = 1 Given this setup, you can prove that given a number m: (me)d = med = m (mod n) What does this do for us, though?

RSA public key encryption p: prime number q: prime number n = pq ϕ(n) = (p-1)(q-1) d: 0 < d < n and gcd(d,ϕ(n)) = 1 e: de mod ϕ(n) = 1 private key public key (d, n) (e, n)

RSA encryption/decryption private key public key (d, n) (e, n) You have a number m that you want to send encrypted encrypt(m) = me mod n (uses the public key) How does this encrypt the message?

RSA encryption/decryption private key public key (d, n) (e, n) You have a number m that you want to send encrypted encrypt(m) = me mod n (uses the public key) Maps m onto some number in the range 0 to n-1 If you vary e, it will map to a different number Therefore, unless you know d, it’s hard to know original m was after the transformation

RSA encryption/decryption private key public key (d, n) (e, n) You have a number m that you want to send encrypted encrypt(m) = me mod n (uses the public key) decrypt(z) = zd mod n (uses the private key) Does this work?

RSA encryption/decryption encrypt(m) = me mod n decrypt(z) = zd mod n decrypt(z) = decrypt(me mod n) z is some encrypted message = (me mod n)d mod n definition of decrypt = (me)d mod n modular arithmetic = m mod n (me)d = med = m (mod n) Did we get the original message?

RSA encryption/decryption encrypt(m) = me mod n decrypt(z) = zd mod n decrypt(z) = decrypt(me mod n) z is some encrypted message = (me mod n)d mod n definition of decrypt = (me)d mod n modular arithmetic = m mod n (me)d = med = m (mod n) If 0 ≤ m < n, yes!

RSA encryption: an example p: prime number q: prime number n = pq ϕ(n) = (p-1)(q-1) d: 0 < d < n and gcd(d,ϕ(n)) = 1 e: de mod ϕ(n) = 1 p = 3 q = 13 n = ? ϕ(n) = ? d = ? e = ?

RSA encryption: an example p: prime number q: prime number n = pq ϕ(n) = (p-1)(q-1) d: 0 < d < n and gcd(d,ϕ(n)) = 1 e: de mod ϕ(n) = 1 p = 3 q = 13 n = ?

RSA encryption: an example p: prime number q: prime number n = pq ϕ(n) = (p-1)(q-1) d: 0 < d < n and gcd(d,ϕ(n)) = 1 e: de mod ϕ(n) = 1 p = 3 q = 13 n = 3*13 = 39

RSA encryption: an example p: prime number q: prime number n = pq ϕ(n) = (p-1)(q-1) d: 0 < d < n and gcd(d,ϕ(n)) = 1 e: de mod ϕ(n) = 1 p = 3 q = 13 n = 39 ϕ(n) = ?

RSA encryption: an example p: prime number q: prime number n = pq ϕ(n) = (p-1)(q-1) d: 0 < d < n and gcd(d,ϕ(n)) = 1 e: de mod ϕ(n) = 1 p = 3 q = 13 n = 39 ϕ(n) = 2*12 = 24

RSA encryption: an example p: prime number q: prime number n = pq ϕ(n) = (p-1)(q-1) d: 0 < d < n and gcd(d,ϕ(n)) = 1 e: de mod ϕ(n) = 1 p = 3 q = 13 n = 39 ϕ(n) = 24 d = ? e = ?

RSA encryption: an example p: prime number q: prime number n = pq ϕ(n) = (p-1)(q-1) d: 0 < d < n and gcd(d,ϕ(n)) = 1 e: de mod ϕ(n) = 1 p = 3 q = 13 n = 39 ϕ(n) = 24 d = 5 e = 5

RSA encryption: an example encrypt(m) = me mod n n = 39 d = 5 e = 5 decrypt(z) = zd mod n encrypt(10) = ?

RSA encryption: an example encrypt(m) = me mod n n = 39 d = 5 e = 5 decrypt(z) = zd mod n encrypt(10) = 105 mod 39 = 4

RSA encryption: an example encrypt(m) = me mod n n = 39 d = 5 e = 5 decrypt(z) = zd mod n encrypt(10) = 105 mod 39 = 4 decrypt(4) = ?

RSA encryption: an example encrypt(m) = me mod n n = 39 d = 5 e = 5 decrypt(z) = zd mod n encrypt(10) = 105 mod 39 = 4 decrypt(4) = 45 mod 39 = 10

RSA encryption: an example encrypt(m) = me mod n n = 39 d = 5 e = 5 decrypt(z) = zd mod n encrypt(2) = ?

RSA encryption: an example encrypt(m) = me mod n n = 39 d = 5 e = 5 decrypt(z) = zd mod n encrypt(2) = 25 mod 39 = 32 mod 39 = 32 decrypt(32) = ?

RSA encryption: an example encrypt(m) = me mod n n = 39 d = 5 e = 5 decrypt(z) = zd mod n encrypt(2) = 25 mod 39 = 32 mod 39 = 32 decrypt(32) = 325 mod 39 = 2

RSA encryption in practice For RSA to work: 0 ≤ m < n What if our message isn’t a number? What if our message is a number that’s larger than n?

RSA encryption in practice For RSA to work: 0 ≤ m < n What if our message isn’t a number? We can always convert the message into a number (remember everything is stored in binary already somewhere!) What if our message is a number that’s larger than n? Break it into m sized chunks and encrypt/decrypt those chunks

RSA encryption in practice encrypt(“I like bananas”) = 0101100101011100 … encode as a binary string (i.e. number) 4, 15, 6, 2, 22, … break into multiple < n size numbers 17, 1, 43, 15, 12, … encrypt each number

RSA encryption in practice decrypt((17, 1, 43, 15, 12, …)) = 4, 15, 6, 2, 22, … decrypt each number 0101100101011100 … put back together “I like bananas” turn back into a string (or whatever the original message was) Often encrypt and decrypt just assume sequences of bits and the interpretation is done outside